Physics 11th Newton's Laws of Motion

Newton's Laws of Motion

(1) Introduction

• While studying kinematics,we have already studied about the position ,displacement and acceleration of a moving particle
• Here in this chapter,we would take our understanding one step further to learn abouts origins of accleration or force
• Here we will specifically consider the cause behind the moving objects i.e what causes the objects to move
• Thus we will learn the theory of motion based on the ideas of mass and force and the laws connecting these physical concepts to the kinematics quantities
• So we will begin by stating the Newton's law's of motion which are of critical importance in classical mechanics
• Laws of motion as stated by Newtons in his principle are
  (i) Every body continues in its state of rest or uniform motion in straight line ,unlesss compelled to change that state by force imposed upon it
  (ii) Change of motion of an objects is proportional to the force acting on it and is made in the direction of the straight line along the direction of force
  (iii) To every action there is always an equal and opposite reaction

(2) Force

• Concept of force is central to all of physics whether it is classical physics,nuclear physics,quantum physics or any other form of physics
• So what is force? when we push or pull anybody we are said to exert force on the body
• Push or pull applied on a body does not exactly define the force in general.We can define force as an influence causing a body at rest or moving with constant velocity to undergo an accleration
• There are many ways in which one body can exert force on another body
  Few examples are given below
  (a)Stretched springs exerts force on the bodies attached to its ends
  (b)Compressed air in a container exerts force on the walls of the container
  (c) Force can be used to deform a flexible object
  There are lots of examples you could find looking around yourself
• Force of gravitational attraction exerted by earth is a kind of force that acts on every physical body on the earth and is called the weight of the body
• Mechanical and gravitation forces are not the only forces present infact all the forces in Universe are based on four fundamental forces
  (i) Strong and weak forces: These are forces at very short distance (10^-05 m) and are responsible for interaction between neutrons and proton in atomic nucleus
  (ii) Electromagnetic forces: EM force acts between electric charges
  (iii) Gravitational force -it acts between the masses
• In mechanics we will only study about the mechanical and gravitational forces
• Force is a vector quatity and it needs both the magnitude as well as direction for its complete description
• SI unit of force is Newton (N) and CGS unit is dyne where
  1 dyne= 10⁰⁵ N

(3) Newton's First law of Motion

• We have already stated Newton's First law of motion which says that a body would continue to be in state of rest or continue to move with constant velocity unless acted upon by a net external force
• Here the net external force on the body is the vector sum of all the extenal forces acting on the body
• When the body at rest or in a state of motion with uniform velocity then in both the cases acceleration is zero.This implies that
  a=0 for F=0
• When net forces i.e vector sum of all the forces acting on the body is zero.the body is said to be in equilbrium .When rotational motion is involved <,net torque on body should also be zero i.e their is no change in either translational or rotational motion
• Since forces can be combined according to the rules of vector addition.Thus for a body to be in equilibrium
  R=ΣF=0
  or in component form
  ΣF_x=0
  ΣF_y=0
  These are the condition for the body in translational equilbrium
• We will discuss about rotational equilibrium while studying torque and rotational motion
• Thus Newton's First law of motion quantitatively defines the concept of force as a influence that changes the state of motion of the body
• It does not say anything about what has to be done to keep object moving that is once the body gains motion by the application of force would it always
• According to first law if we completely eliminates frictional forces,no forward force at all would be required to keep an object ( say a block on table)

(4) Inertia and Mass

• From First law of motion an object at rest would not move unless it is acted upon by a force
• This inherent property of objects to remain at rest unless acted upon by a force is called intertia rest
• Now consider the case of an object moving with uniform velocity along the straight line .Again from Newton's law it would continue to move with uniform
• This inherent property by virtue of which a body in state of uniform motion tend to maintain its uniform motion is called inertia of motion
• Combining these two statements 'The property of an object to remain in state of rest or uniform rectilinear motion unless acted upon by a force is called inertia
• Mass of any body is the measure of inertia .For example if we apply equal amount of force on two objects of different mass (say m₁ and
  m₂ such that m₁ > m₂ ) Tnen acceleration of both the object would be different (i.e , a₁ < a₂ )
• Acceleration of object having larger mass would be lesser then the acceleration of object having smaller mass
• Thus larger the mass of the body ,smaller would be the acceleration and larger would be the inertia
• Newton's first law of motion revealing this fundamental property of matter i.e inertia is also known as law of inertia

(5) Newton's second law of motion

• Newton's first law of motion qualitatively defines the concept of force and the principle of inertia
• For an body at rest,application of force causes a changed in its existing state and application of force on a body moving with uniform velocity would give the body under consideration as acceleration
• Newton's second law of motion is a relation between force and acceleration
• Newton's second law of motion says that
  " The net force on a body is equal to the product of mass and acceleration of the body"
  Mathematically
  F_net=ma                               (1)
  Where F_net is the vector sum of all the forces acting on the body
• Above equation -(1) can be resolved along x,y and z components .Thus in component form
  F_netx=ma_x
  F_nety=ma_y
  F_netz=ma_z
• Component of accleration along a given axis is caused only by the net component of force along that axis only not by the components of force along other
• Newton's second law of motion is completely consistent with newton first law of motion as from equation (1) F=0 implies that a=0
• For a body moving under the influence of force,acceleration at any instant is determined by the force at that instant not by the prev ious motion of the
• Newton's second law of motion is strictly applicable to a single particle .In case of rigid bodies or system of particles,it refers to total external forces acting on the system excluding the internal forces in the system.

(6)Newton's third law of motion

• Statement of newton's third law of motion is " To every action there is always an equal and opposite reaction".
• Thus,whenever a body exerts force on another then another object exert an equal force on previous body but in opposite direction
• Force example motion of rocket depends on the third law of motion i.e, action and reaction .Rocket exerts action force on gas jet in backward direction
• Force of action and reaction acts on different objects i.e,
  Force object 1 exerts on object 2= Force object 2 exerts on object 1
  i.e,
  F₁₂=-F₂₁
  Action=-(Reaction)
• According to newtonian mechanics force is always a mutual interaction between the bodies and force always occurs in pairs
• Equal and opposite mutual forces between two bodies is the basic idae between Newton's third law of motion
• While considering a system of particles ,internal force always cancel away in pairs i.e consider two particles in a body if F₁₂ and
  F₂₁ are internal forces between partcile system 1 and 2 then they add up to give a null internal force.Same way internal forces for the particles

(7) Applying Newton's law of motion

• Newton's law of motion ,we studied in earliar topics are the foundation of mechanics and now we look forward to solve problems in mechanics
• In general,we deal with mechanical systems consisting of different objects exerting force on each other
• While solving a problem choose any part of the assembly and apply the laws of motion to that part including all the forces on the choosen part of the assembly due to remaining parts of the assembly
• Following steps can be followed while solving the problems in mechanics
  1)Read the problem carefully
  2) Draw a schematic diagrom showing parts of the assembly for example it may be a single particle or two blocks connected to string going over pulley etc
  3) Identify the object of prime interest and make a list of all the forces acting on the concerned object due to all other objects of the assembly and exclude the force applied by the object of prime interest on the other parts of the assembly
  4) Indicate the forces acting on the concerned object with arrow and Lable each force for example tension on the object under consideration can be labelled by letter T
  5) Draw a free body diagram of the object of interest based on the labelled picture.Free body diagram for the object under consideration shows all the forces exerted on this object by the other bodies.Do not forget to consider weight W=mg of the body while labelling the forces acting on the body
  6) If additional objects are involved draw seperate free body diagrma for them also
  7)Resolve the rectangular components of all the forces acting on the body
  8) Write Newton second law of equation for the body and solve them to find out the unknowm quantities
  9) Do not forget to employ Newton's third law of motion for action reaction pair which results in null resultant force
• Following solved example would clearly illustrate how to apply Newton's laws of motion follwowing the above given procedure

Solved Example :

Question:
A horizontal forces of magnitude 500N pulls two blocks of masses m₁=10 kg and m₂=20 kg which are connected by the light inextensible string and lying on the horizontal frictionless surface.Find the tension in the strings and acceleration of each mass when forces is applied on mass
m₂? Solution:
Given that force is applied on the block m₂ as shown in the figure below

Let T be the tension in the string and a be the accleration of each mass .Now we will draw free body diagrams for each masses


Weights of the blocks m₁g and m₂g are balanced by their normal reaction R₁ and R₂ respectivley.The equations of motion of the two massed are found using Newton's second law of motion
m₁a=T ...............................(1)
m₂a=F-T ............................(2)

Dividing 1 by 2
we get
T=m₁F/(m₁+m₂)
Substituting the given values
T=166.7 N
Using value of T in equation 1 ,we find
a=16.67 m/s²
Above sample problem shows how to solve a typical mechanics problem.Similarly by adopting given procedure we can solve other such problems 

(8) Inertial frame of Reference

• Inertial frame of refrences are those frames of reference in which newton's first and second law of motion is always hold true
• A frame of reference in which Newton's law are not valid is called non-inertial frame of refrence
• In an inertial frame if a body is not acted by external force ,it continues to be in state of rest or uniform translatory motion.Thus in an inertial frame if the body is not acted upon by an external force then accleration would be zero
  a=(d²r/dt²)=0
• If a frame is inertial frame ,then all those frames which are moving with constant velocity relative to the previous frame are also inertial frames
• Inertial frame of refrence are necessarly unaccelerated frames because if the frame is acclerated the particle moving with uniform velocity will appear

(9) Fictitious ( or Pseudo ) Forces

• We already know about Non-inertial frame of reference .All the accelrated and rotatig frame of reference are non-inertial frame of refrence
• Consider an interial frame of reference S and let S^' be any other frame moving with accleration w.r.t to frame S as shown in the below figure
  
   
• Now if no external forces are acting on particle P .Then its acceleration would be zero in Frame S but in frame S^',an observer will find an acceleration -a₀ acting on the particle.
• The observer force on particle P of mass m in Frame S^' is -ma₀
• But in reality no such force is acting on the particle and particle appears to be accelerated in this non-inertial frame of reference.Such one force is known as Pseudo or Fictitious Force. Hence Pseudo Force on particle is
  F_P=-ma₀
• Now if we apply F_i on the particle and a_i is the observed accleration of particle in S frame(Inertial frame) The according to Newton's law
  F_i=ma_i
• For calculating net force in accelerated frame consider both the frames S and S^' coincid at time t=0 .After time t let r_iand r_n be the position vector of the particle in frame S and S^' respectively
  The relation between r_i and r_n is
  r_i =r_n +(1/2)a₀t² Where a₀ is the acceleration of frame S^' wrt frame S
  Differentiating the equation w.r.t time twice
  a_i=a_n + a₀
  or ma_i-ma₀=ma_n
  => F₀ + F_P=F_N
  This equation gives obesrved force in acclerated frame of reference

Solved Examples

Question 1 Force exerted on a body changes it's

(a) Direction of motion
(b) Momentum
(c) Kinetic energy
(d) All the above

Solution: A body acted upon by a certain force produces acceleration i.e. it undergoes change in it's velocity. hence choice (d) is correct

Question 2 there are two statements
A Rate of change of momentum correspondes to force
B Rate of change of momentum corresponds to Kinetic Energy
Which of the following is correct

(a) A only
(b) B only
(c) Both A and B are correct
(d) Both A and B are wrong

Solution 

F=dp/dt so a is correct


Question 3 A truck and car are moving on a plane road with same kinetic Energy. They are brought to rest by application of brakes which provide equal retarding forces. Which of the following statements is true.

(a) Distance travelled by truck is shorter then car before comming to rest
(b) Distance travelled by car is shorter then truck before comming to rest
(c) Distance travelled depends on individual velocity of both the vehicles
(d) Both will travel same distance before comming to rest

Solution 3
By work energy therorm we know
KE_f-KE_i=Workdone

So in this case
Intial KE=retarting force * distance travelled

Since kinectic energy is same and retarding force is equal

Distance travelled will be same


Question 4 A block A of mass m₁ is released from top of smooth inclined plane and it slides down the plane. Another block of mass m₂ such that m₂ > m₁is dropped from the same point and falls vertically downwards. Which one of the following statements will be true if the friction offered by air is negligible?

(a) Both blocks will reach ground at same time
(b) Both blocks will reach ground with the same speed
(c) speed of both the blocks when they reach ground will depend on their masses
(d) Block A reaches ground brfore block B

Solution 4

since velocity of block when it reaches the ground is given by v=(2gh)^1/2 and it is independent of the mass the correct choice will be (b).


Question 5 Which of the following observer is/are non interial
a.A child revolving in the merry ground
b. A driver in a train moving with constant velocity v
c. A passenger in a bus which is slowing down to a stop
d None of these

Solution 5

Since centripetal acceleration is present in merry round it is non interial frame of refrence

Since bus is slowing down to a stop,it means it is experiencing a retardation ,so it is non interial frame of refrence


Question 6 there are two statements
A Newtons first law in valid from the pilot in an aircraft which is taking off
B Newtons first law in valid from the observer in a train moving with constant velocity
Which of the following is correct

(a) A only
(b) B only
(c) Both A and B are correct
(d) Both A and B are wrong

Solution 6

We know that Newtons law is invalid in Non interial frame of refrence, now since acceleration is present in Ist case..So answer is B


Question 7A IITJEE text book of mass M rests flat on a horizontal table of mass m placed on the ground.Let R_X->Y be the constant force exerted by the body x on body Y.According to Newton third law,which of the following is an action-reaction pair of forces?

a. (M+m)g and R_table->book
b. R_{ground->table} and mg+R_book->table
c. R_{ground->table} and R_{table->ground}
d Mg and R_table->book

Solution 7

Action reaction pair acts on diffrent body and always are in oppsite direction.We will have to draw free body diagram for each part in this case.



Question 8 Choose the correct alternative
a. The acceleration of a moving particle is always in the direction of its velocity
b. Velocity and acceleraion vectors of a moving particle may have any angle between 0 and 360 between them
c.Velocity and acceleraion vectors of a moving particle may have any angle between 0 and 180 between them
d. If the acceleration vector is always perpendicular to the velocity vector of a moving particle,the velocity vector does not change

Solution 8

In circular motion,acceleration vector is perpendicular to velocity and velocity is also changing.

Now acceleration and velocity can have any angle between

Question 9 there are two statements
A Newtons first law in valid from the pilot in an aircraft which is taking off
B Newtons first law in valid from the observer in a train moving with constant velocity
Which of the following is correct

(a) A only
(b) B only
(c) Both A and B are correct
(d) Both A and B are wrong

Solution 9 Newtons law is valid from Intertial frame of refrence
when the plane is taking off,it has acceleration so it can not be intertial frame of refrence
When the observer is train moving with constant velocity,so no acceleration is invloved.So it is Intertial frame.Hence Newton law is valid

Question 10.A body of mass 5 kg starts from the origin with an intial velocity u=30i+40j m/sec.If the constant force acts on the body F=-(i+5j) N.The time in which y component of the velocity becomes zero is
a. 5 sec
b. 80 sec
c. 20 sec
d. 40 sec

Solution 10.

Force=-(i+5j) N
Mass =5 Kg
intial velocity u=30i+40j m/sec

Acceleration:-
a=F/m=-(i+5j)/5=-.2i-j

Y component of intial velocity is 40 m/s
Acceleration in that direction -1 m/s²

So
v=u+at
0=40-1t
or t=40 sec

Hence d is correct


Question 11.A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle.The motion of the particle takes place in a plane.it follows that
a. velocity is constant
b. accelerattion is constant
c. KE is constant
d. Particle moves in a circular path

Solution 11
Since force is always perpendicular to the body,workdone by it will be zero
So kinetic energy is constant
It is the case of uniform circular motion
Velocity magnitude is constant but direction is constantly changing
Acceleration is constany but direction is constantly changing

Hence c and d are constant


Question 12A small mass m is suspended from one end of a vertical string. and then whirled in a horizontal circle at a constant speed v.
Which of the followings is true
a.The strings stays vertical
b.The string becomes inclined to the vertical.
c.There is no force on mass m except its weight
d.The angle of inclination of the string does not depend on the v
e.The centripetal force on m is mg

Solution 12
Since the body is rotating on a horizontal circle,String will be inclined to the horizontal
It is shown in figure

The forces acting on the body are weight and tension in the string
Vertical component of the tension will balance out the weight of the body
Horizontal component will provide the required centripetal force for the circular motion

if T is the tension and W be the weight and θ be the angle of inclination
then
Tcosθ =Wg
And Tsinθ =mv²/R

So angle of inclination depends on the speed of the motion
So only correct answer is b

Question 13.The tension in cable supporting an elevator is equal to th weight of the elevetor.The elevator may be
a. going up with incresing speed
b. going down with increasing speed
c. going up with uniform speed
d. going down with uniform speed.

Solution 13
Since the tension is the string equals weight of the elevator,It means resultant force is zero
So acceleration must be zero
Hence c and d are correct answer


Question 14 A refrence frame attached to the earth
a.is an inertial frame by defination
b.cannot be inertial frame as earth is revolving around the sun
c.is an intertial frame because Newton's law are applicable in this frame
d.cannot be intertial frame because earth is rotating about its own axis

Solution 14
Earth cannot be inertial frame as it is revolving around the sun and also rotating about its axis

Question 15By means of rope ,a body of weight W is moved vertically upward with constant acceleration a .Find the tensile force in the rope
a. W(1+a/g)
b W(1-a/g)
c W
d. none of the above

Solution 15
Forces acting on the body
Tensile force vertically upward=S
Weight of the body vertically downward=W
Body has acceleration a vertically upwards

Net force on the body must provide the acceleration
so S-W=(W/g)a
or S=W(1+a/g)

Physics 11th Friction

Friction

(1) Introduction

• We all know that what would happen if we slide a book kept along a horizontal table .Yes it would first sliding and then finally would come to rest
• This force opposing continous motion of book on the table is called force of friction
• Whenever surface of one object slides over another ,both the bodies exerts force of friction on each other
• Here in this chapter,we will consider frictional forces acting between a pair of surfaces
• Frictional force comes into action whenever two surfaces of two bodies comes in contact with each other(Both the bodies could be moving or rest).Both the bodies exert a force on each other which is basically E.M in nature and this force is a contact force
• Magnitude of contact forces on two bodies are equal but they are opposite in direction
• Contact forces also depend on the nature of the surfaces of the two bodies kept in contact,how both the bodies are moving and all the forces acting on them
• Consider a block resting on the horizontal surface as shown below in the figure ,Now two bodies exert equal and opposite contact force on each other . Let F_c is the contact force both the bodies exert on each other
   
• Direction of the contact force acting on a body need not necessarily be perpendicular to the surface of contact
• Contact force F_c can be resolved into two components.The component of F_c along the normal to the surface is called Normal
  contact force and component parallel to the surface is called friction (f) as shown in fig (b)
• Friction is subdivided into two catagories namely
  1. Static Friction
  2. Kinetic Friction
  Static friction is experienced betweem the non moving surfaces i.e the bodies at rest and kinetic friction or dynamic friction is experienced between the moving surfaces or between the bodies in motion
• We shall now discuss both the static and kinetic friction in details

(2) Static Friction

• We already know that frictional forces can also act between the bodies in contact with each other even if they are not moving and such type frictional force is known as static friction
• Consider a heavy metal block kept on the floor and you are trying very hard to push it to another location and you are not able to slide it even by a centimeter
• Since the block is at rest resultant force on it should be zero.To counter balance the force applied by you ,floor exerts a frictional force on the block s
• Now if you begin to increase the magnitude of force gradually then bloack does not start moving until force applied is greater then a minimum value of force
• This force of static friction must be overcome by the applied force before the body at rest begin to move
• Force of friction is always equal and opposite to the external applied force as long as the body is at rest
• This means that static friction force is a self adjusting force.It adjust its value accordingly with the increase in magnitude of applied force
• This frictional force cannot be unlimited and its value cannot go beyond a maximum value f_ms
• This maximum value of static friction between the two surfaces in contact is known as limiting friction
• Thus magnitude of static friction can not go beyond the magnitude of the limiting friction i.e, f_s <= f_ms
• This limiting friction is proportional to the normal contact force (N) between the bodies i.e,
  f_ms is proportional to N
  f_ms=μ_sN
  Where N =Normal contact force
  μ_s is the proportionality constant known as coefficent of static friction
• Value of coefficent of static friction depends on the material and roughness of the surfaces of bodies in contact
• f_ms is the maximum value that force of static friction acting between two bodies can reach
• The actual force of static friction can be equal to zero or less then f_ms and its value depends on the force applied on the body thus
  f_s <= f_ms=μ_sN

(3) Kinetic or dynamic friction

• We already know that static frictional force exists between two surfaces in contact before there is relative motion between two surfaces
• When applied force becomes greater then limiting friction force,then motion of the body starts and kinetic friction comes into existence
• Thus when bodies in contact moves relative to each other then friction devloped between them is called kinetic or dynamic friction
• Consider the figure below in which a block of mass m is kept over the surface S
   
• Initially the block is at rest,now we push the block on the surface such that it begins to move or start sliding on the surface
• When the block is sliding over the surface each body exerts a frictional force on the other side parallel to the surface in contact
• The force of friction acting on the block B due to surface S is along the direction opposite to the motion of B with respect to S.Thus force of kinectic friction opposes the relative motion between the bodies in motion
• The frictional force acting on the block is along the direction towards the left and an equal force acts on the surface S directed towards the rights
• Thus we can say that sliding friction on a body B against surface S is opposite to the velocity of the body B with respect to S
• Force of kinetic friction is denoted by f_k and its magnitude is always less then the magnitude of limting friction i.e maximum static friction
  i.e f_k < f_ms
• kinetic friction force is proportional to the normal contact force between the surfaces i.e
  f_k is proportional to N
  => f_k =μ_kN
  Where N =Normal contact force
  μ_k is the proportionality constant known as coefficent of kinetic friction and its value depends on the nature of the two surfaces in contact
• f_k < f_ms, this inequality shows that force required to start the motion is greater then the force required to maintail the further motion of the body

(4) Rolling Friction

• Consider a situation of the ring or a sphere rolling without slipping over a horizontal plane.In this case there is only one point of contact between the body and the plane
• The frictional forces developed between two surfaces in case described above is called rolling friction
• Rolling friction developes between two surfaces when one body rolls over the surface of another body
• We know that it is very difficult to pull a heavy metal box on a rough surface and if we attah four metal wheel to the box it becomes easiar to move the
• Thus resistance offered by the surface during rolling is relatively less than offered during sliding friction
• This is because while rolling surfaces in contact do not rub each other
• Rolling friction is negligible in comparision to the kinetic and static friction which are present simlutanously
• In many parts of the machine where this type of friction is undesirable ball bearings(small steel balls) are generally kept between the rotating parts of

(5) Angle of friction

• It is the angle which the normal force N makes with the contact force when the equlibrium is limiting i.e when the condition of maximum static friction
• Consider a block of mass m resting on a horizontal surface .Weight W=mg of this block is balanced by the normal force (N) of reaction as shown below in the figure.
  
  
  Mathematically
  W=mg=-N
• if F is the contact force which each body exerts on the another body then this contact force can be resolved into two components
• Perpendicular Components which is normal force N
• Components parallel to the contact surface known as friction which is usually denoted by the f_s,f_k or f_ms
• Angle BCD respresented by λ is angle of friction
  Since N and f_ms are components of the contact force F,so we can write
  Fcosλ=N
  and Fsinλ=f_ms
  or we can say
  tanλ=f_ms/N
  Also from the defination of static friction we know that
  f_ms=μ_sN So tan λ=μ_s
  Thus we can conclude that coefficient of static friction is equal to the tangent of the angle of friction

(6) Methods to Reduce Friction

• Friction can be reduced by the numbers of methods.Some of them are listed below
  (a)Lubrication: when a lubricant is applied between the two surfaces in contact then a thin layer of lubricant is formed between the two surfaces resulting in reduction in friction.Example of Lubricant Grease oil,graphite,compressed air
  (b)Polishing : Irregularities of a surface can be reduced by polishing the surface smooth which results in reduction in friction
  (c) Friction between the two surfaces can be reduced by using ball bearing between the two surfaces.Also sliding friction can be converted into rolling friction which is much less then sliding friction

Solved examples

Question 1 A block of Mass M is moving with a velocity v on straight surface.What is the shortest distance and shortest time in which the block can be stopped if μ is coefficent of friction
a.v²/2μg,v/μg
b. v²/μg,v/μg
c.v²/2Mg,v/μg
d none of the above

Solution 1
Force of friction opposes the motion
Force of friction=μN=μmg
Therefore retardation =μmg/m=μg

From v²=u²+2as
or
S=v²/2μg

from v=u+at
or t=v/μg


Question 2A horizontal force of F N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is μ. The weight of the block is
a.μF
b. F(1+μ)
c. F/μ
d none of these

Solution 2

Let W be the weight
Reaction force=F
Weight downward=W
weight Upward=frictional force=μr=μF

For no movement
weight Upward=Weight downward
W=μF

Question 3.A body is sliding down a rough inclined plane of angle of inclination θ for which coefficent of friction varies with distance y as μ(y)=Ky where K is constant.Here y is the distance moved by the body down the plane.The net force on the body is zero at A.Find the value of constant K
a. tanθ/A
b. Acotθ
c. cottanθ/A
d. Atantanθ

b>Solution 3


The downward force=mgsinθ
The upward force=μmgcosθ

Net force
f(y)=mgsinθ-μmgcosθ
=mg(sinθ-kycosθ)

at y=A, f(y)=0
0=sinθ-kAcosθ
or K=tanθ/A


Question 4A given object takes n times as much times to slide down a 45 rough incline as its takes to slide down a perfectly smooth 45 incline.The coefficent f kinectic friction between the objects and incline is gievn by.
a. 1/(1-n²)
b.1-1/n²
c. √1/(1-n²)
d. √(1-1/n²)

Solution 4

let μ be the coefficient of friction

Acceleration is smooth 45 inclined plane=gsinθ=g/(2)sup>1/2
Acceeration in friction 45 inclined plane=gsinθ-μqcosθ=[g/(2)sup>1/2(1-μ)

Now s=ut+(1/2)at²
or 2s=at² as =0

For smooth plane
2s=g/(2)sup>1/2t² ---(1)

for friction plane
2s=[g/(2)sup>1/2(1-μ)(nt)² -(2)

so (1-μ)n²=1
or μ=1-1/n²


Question 5A uniform chain of length L is lying on the horizontal surface of a table.If the coefficent of friction between the chain and the table top is μ. what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on table.?
a.L/(1+μ)
b. μL/(1+μ)
c. L/(1-μ)
d. μL/(1-μ)

Solution 5

Let m be the mass per unit length

Let L be the full length and l be the length of chain hanging

So net force downwards=mlg

Net frictional force in opposite direction=μm(L-l)g

Now mlg=μm(L-l)g

or l=μ(L-l)
or l=μL/(1+μ)



Question 6The coefficent of static and kinectic friction between a body and the surface are .75 and .50 respectively.A force is applied to the body to make it just slide with a constant acceleration which is
a. g/4
b g/2
c. 3g/4
d g

Solution 6

Minimum force with which body will just move=μ_smg

After the body start moving Frictional force becomes =μ_kmg

So ma=μ_smg-μ_kmg
or a=g/4


Question 7 A block of mAss M is moving with a velocity v on straight surface.What is the shortest distance and shortest time in which the block can be stopped if μ is coefficent of friction
a.v²/2μg,v/μg
b. v²/μg,v/μg
c.v²/2Mg,v/μg
d none of the above

Solution 7
Force of friction opposes the motion
Force of friction=μN=μmg
Therefore retardation =μmg/m=μg

From v²=u²+2as
or
S=v²/2μg

from v=u+at
or t=v/μg



Question 8A horizontal force of F N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is μ. The weight of the block is
a.μF
b. F(1+μ)
c. F/μ
d none of these

Solution 8

Let W be the weight
Reaction force=F
Weight downward=W
weight Upward=frictional force=μr=μF

For no movement
weight Upward=Weight downward
W=μF

Physics 11th Work energy and Power

Work energy and Power

(1) Introduction

• In our everyday life we use terms like work and enegy.
• Term work is generally used in context to any kind of activity requiring physical or mental effort.
• But this is not the way how we define work done in physics.
• When we push or pull a heavy load or lift it above the floor then we are doing work, but a man carrying heavy load and standing still is not doing any work according to scientific definiton of work.
• Another term we often use is energy. Energy is usually associated with work done in the sence that a person feeling very energetic is capable of doing lot of work.
• This way energy defined to be as capacity of doing work.
• There are many forms of energy like chemical energy, mechanical energy, electrical energy, heat energy etc. These forms of energies can be used in number of ways.
• One form of energy can be converted into another form of energy.
• In this chapter we will study about work, relation between work and energy, conservation of energy etc. 

(2) Work

• We already know that work is said to be done when a force produces motion.
• Work done is defined in such a way that it involves both force applied on the body and the displacement of the body.
• Consider a block placed on a frictionless horizontal floor. This block is acted upon by a constant force F. Action of this force is to move the body through a distance d in a straight line in the direction of force.
• Now , work done by this force is equal to the product of the magnitude of applied force and the distance through which the body moves. Mathematically,
  W=Fd                         (1)
  where, F=|F|
• Consider the figure given below
  
  
• In this case force acting on the block is constant but the direction of force and direction of displacement caused by this force is different. Here force F acts at an angle θ to the displacement d
• Effective component of force along the direction of displacement is Fcosθ and this component of force is responsible for the displacement of the block in the given direction.
• Thus , work done by the force F in in displacing the body thruugh displacement d is
  W=(|F|cosθ)|d|          (2)
  In equation 2 work done is defined as the product of magnitude of displacement d and the component of force in the direction of displacement.
• We know that the scalar product of two vectors A and B where A makes an angle θ with B is given by
  
  
  
  A.B=|A| |B|cosθ
• Comparing equation 2 with definition of scalar products work done can be written as
  W=F.d               (3)
• Now consider two special cases:-
  (i) When angle θ=0 i.e., force is in the same direction as of displacement, then from equation (2)
  W=|F| |d|
  this is the same result as of equation 1
  (ii) when angle θ=90 i.e., direction of force is perpandicular to that of displacement, then from equation (2)
  W=(|F|cos90)|d|=0
  that is force applies has no component along the displacement and hence force does not do any work on the body.
• Work done by a force on a body can be positive, negative and zero i.e.,
  (a) Work done is positive :- Force is in the same direction as displacement for example work done by force for pushing a block of mass m
  (b) Work done is negative :- Force is opposite to the displacement for example when a body slides on horisontal surface work done by frictional forces on body is negative as frictional force always acts opposite to displacement of the body.
  (c) Work done is zero :- Force is at rigt angle to the displacement for example work of a centripetal force on a body moving in a circle.
• Unit of work done in any system of units is equal to the unit of force multiplied by the unit of distance. In SI system unit of work is 1Nm and is given a name Joule(J). Thus,
  1J=1Nm
  In CGS system unit is erg
  1erg=1 dyne-cm
  and 1 erg=10^-7J
• When more then one forces acts on a body then work done by each force should be calculated separately and added togather

(3) Work done by variable force

• So far we have defined work done by a force which is constant in both magnitude and direction.
• However, work can be done by forces that varies in magnitude and direction during the displacement of the body on which it acts.
• For simplicity consider the direction of force acting on the body to be along x-axis also consider the force F(x) is some known function of position x
• Now total displacement or path of the body can be decomposed into number of small intervals Δx such that with in each interval force F(x) can be considered to be approximately constant as shown below in the figure
  
  
• Workdone in moving the body from x₁ to x₂ is given by
  ΔW=F(x₁)Δx₁ where Δx₁=x₂-x₁
• Total workdone in moving the body from point A to point B
  W=F(x₁)Δx₁ +F(x₂)Δx₂ +F(x₃)Δx₃ .....+F(x_n)Δx_n
  W=Σ F(x_i)Δx_i where i=1 to i=n           (4)
  Where Σ is the symbol of summation
• Summation in equation 4 is equal to shaded area in figure 3(a) .More accuracy of results can be obtained by making these interval infintesimally smaller
• We get the exact value of workdone by making each interval so much small such that #916;x-> 0 which means curved path being decomposed into infinte number of line segment i.e
  W= Lim_{Δx-> 0}Σ F(x_i)Δx_i           (5)
  W =∫ F(x)dx with in the limits x_A and x_B
  Where ∫ F(x)dx is the intregal of F(x) w.r.t x between the limits x_A and x_B and intregal can be evaluated using methods of calculus if F(x) of x is known
• Instead of x if the force also acts along y and z axis i.e direction of force keeps on changing then workdone bu such a force is given by
  
  W=∫ F(r)dr with in the limits r_A and r_B           (6)
  Where F(r)=F(x)i +F(y)j+ F(z)k
  and dr=dxi +dyj+ dzk
  Here F(x),F(y) and F(z) are rectangular components of the force along x ,y and z axis .Similary dx,dy and dz are rectangular components of displacement along
  x ,y and z axis

(4) Mechanical Energy

• we already have an idea that energy is associated closely with work and we have defined energy of a body as the capacity of the body to do work
• In dynamics body can do work either due to its motion ,due to its position or both due to its motion and position
• Ability of a body to do work due to its motion is called kinetic energy for example piston of a locomotive is capable of doing of work
• Ability of a body to do work due to its position or shape is called potential energy For example workdone by a body due to gravity above surface of earth
• Sum of kinetic energy and Potential energy of body is known as its mechanical energy
  Thus
  M.E=K.E+P.E

(5) Kinetic energy

• Kinetic energy is the energy possesed by the body by virtue of its motion
• Body moving with greater velocity would posses greater K.E in comparison of the body moving with slower velocity
• Consider a body of mass m moving under the influenece of constant force F.From newton's second law of motion
  F=ma
  Where a is the acceleration of the body
• If due to this acceleration a,velocity of the body increases from v₁ to v₂ during the displacement d then from equation of motion with constant acceleration we have
  v₂² -v₁²=2ad or
  a=v₂² -v₁²/2d Using this acceleration in Newton's second law of motion
  we have
  F=m(v₂² -v₁²)/2d
  or
  Fd=m(v₂² -v₁²)/2
  or
  Fd=mv₂²/2 -mv₁²/2           (7)
  We know that Fd is the workdone by the force F in moving body through distance d
• In equation(7),quantity on the right hand side mv²/2 is called the kinetic energy of the body
  Thus
  K=mv²/2
• Finally we can define KE of the body as one half of the product of mass of the body and the square of its speed
• Thus we see that quantity (mv²/2) arises purely becuase of the motion of the body
• In equation 7 quantity
  K₂=mv₂²/2
  is the final KE of the body and
  K₁=mv₁²/2
  is the initial KE of the body .Thus equation 7 becomes
  W=K₂-K₁=ΔK           (9)
• Where ΔK is the change in KE.Hence from equation (9) ,we see that workdone by a force on a body is equal to the change in kinetic energy of the body
• Kinetic energy like work is a scalar quantity
• Unit of KE is same as that of work i.e Joule
• If there are number of forces acting on a body then we can find the resultant force ,which is the vector sum of all the forces and then find the workdone on the body
• Again equation (9) is a generalized result relating change in KE of the object and the net workdone on it.This equation can be summerized as
  K_f=K_i+W           (10)
  which says that kinetic energy after net workdone is equal to the KE before net work plus network done.Above statement is also known as work-kinetic energy theorem of particles
• Work energy theorem holds for both positive and negative workdone.if the workdone is positive then final KE increases by amount of the work and if workdone is negative then final KE decreases by the amount of workdone

(6)Potential energy

• Potential energy is the energy stored in the body or a system by virtue of its position in field of force or by its configuration
• Force acting on a body or system can change its PE
• Few examples of bodies possesing PE are given below
  i) Stretched or compressed coiled spring
  ii) Water stored up at a height in the Dam possess PE
  iii) Any object placed above the height H from the surface of the earth posses PE
• Potential energy is denoted by letter U
• In next two topics we would discuss following two example of PE
  i)PE of a body due to gravity above the surface of earth
  ii) PE of the spring when it is compressed orr elongated by the application of some external force

(7) Gravitational PE near the surface of the earth

• All bodies fall towards the earth with a constant acceleration known as acceleration due to gravity
• Consider a body of mass m placed at height h above the surface of the earth
• Now the body begins to fall towards the surface of the earth and at any time t ,it is at height h^' (h^' < h) above the surface of the earth
• During the fall of the body towards the earth a constant force F=mg acts on the body where direction of force is towards the earth
• Workdone by the constant force of gravity is
  W=F.d
  W=mg(h-h^')
  W=mgh-mgh^' -(12)
  From equation (12) ,we can clearly see that workdone depends on tbe difference in height or position
• So a potential energy or more accurately gravitational potential energy can be associated with the body such that
  U=mgh
  Where h is the height of the body from the refrence point
• if
  U_i=mgh U_f=mgh^' or
  W=U_i-U_f=-ΔU           (13)
  The potential energy is greater at height h and smaller at lower height h
• We can choose any position of the object and fix it zero gravitational PE level.PE at height above this level would be mgh
• The position of zero PE is choosen according to the convenience of the problem and generally earth surface is choosen as position of zero PE
• One important point to note is that equation (13) is valid even for the case when object object of mass m is thrown vertically upwards from height h to h^'( h^' > h)
• Points to keep in mind
  i) If the body of mass m is thrown upwards a height h above the zero refrence level then its PE increase by an amount mgh
  ii) If the body of mass falls vertically downwards through a height h ,the PE of body decreases by the amount mgh

(8) Conversion of Gravitational PE to KE

• Consider a object of mass placed at height H above the surface of the earth
• By virtue of its position the object possess PE equal to mgH
• When this objects falls it begins to accelerate towards earth surface with acceleration equal to acceleration due to gravity g
• When object accelerates it gathers speed and hence gains kinetic energy at the expense of gravitational PE
• To Analyze this conversion of PE in KE consider the figure given below
  
• Now at any given height h,PE of the object is given by the U=mgh and at this height speed of the object is given by v
• Change in KE between two height h₁ and h₂ would be equal to the workdone by the downward force
  mv₁²/2 -mv₂²/2 =mgd           (14)
  Where d is the distance between two heights .here one important thing to note is that both force and are in the same downwards direction
• Corresponding change in Potential energy
  U₁-U₂=mg(h₁-h₂)=-mgd           (15)
  Here the negative sign indicates the decrease in PE whch is exacty equal to the gain in KE
• From equation (14) and (15)
  mv₁²/2 -mv₂²/2=U₂-U₁
  or KE₁ + PE₁=KE₂ + PE₂
• Thus totat energy of the system KE and gravitational PE is conserved .There is a conversion of one form of energy into another during the motion but sum remains same
• This can be proved as follows
  At height H,velocity of the object is zero
  ie
  at H=H ,v=0 => KE=0
  So KE+PE=mgH
  at H=h ,v=v => KE=mv²/2=mg(H-h) and U=mgh
  So KE+PE==mgH
  at H=0 ,v=v_max => KE=mv_max²/2=mgH
  So KE+PE=mgH
• This shows that total energy of the system is always conserved becuase total energy KE+PE=mgH ,which remains same as plotted in the figure given below
  
  
• We know conclude that energy can neither be produced nor be destroyed .It can only be transformed from one form to another

(9) PE of the spring

• To study this ,consider an electric spring of negligibly small mass .One end of the spring is attached to the rigid wall and another end of spring is attached to a block of mass m which can move on smooth frictionless horizozntal surface
• Consider the figure given below
  
  
  
  Here unstrectched or un compressed position of the spring is taken at x=0
• We now take the block from its unstrecthed position to a point P by stretching the spring
• At this point P restorinng force is exerted by the spring on the block trying it bring it back to the equlibrium position.
• Similar restoring force developes in the spring when we try to compress it
• For an ideal spring ,this restoring force F is proportional to displacement x and direction of restoring force is opposite to that displacement
• Thus force and displacement are related as
  F α x
  or F= - kx           (16)
  where K is called the spring constant and this equation (16) is known as Hook's law.negative sign indicates that force oppose the motion of the block along x
• To stetch a spring we need to apply the external force which should be equal in magnitude and opposite to the direction of the restoring force mentioned above i.e for stretching the spring
  F_ext=Kx Similary for compressing the spring
  - F_ext= - Kx
  or F_ext=Kx (both F and x are being negative)
• Workdone in both elongation and compression of spring is stored in the spring as its PE which can be easily calculated
• If the spring is stretched through a distance x from its equilibrium position x=0 then
  W=∫F_extdx
  Since both F_ext and dx have same direction Now
  W=∫Kxdx
  On integrating with in the limits x=0 to x=x
  We have
  W=Kx²/2           (17)
• This workdone is positive as force is towards the right and spring also moves towards the right
• Same amount of external is done on the spring when it is compressed through a distance x
• Workdone as calculated in equation (17) is stored as PE of the spring.Therefore
  U=Kx²/2           (18)

(10)Conservation of energy of spring mass system

• Again consider the figure (6) where we have stretched the spring to a distance x₀ from its equlibrium position x=0 
• If we release the block then speed of the block begin to increase from zero and reach its maxmimum value at x=0 (equilbrium position)
• Assuming that there is no dissaption of energy due to air resistance and frictional forces ,whole of the PE is coverted in KE
• This means gain in KE is exactly equal to the loss in PE
• After reaching the equilbrium position at x=0 ,the block then crosses the equilibrium position ,it speed begin to decreass until it reaches the point x=-x₀ 
• This would happen if whole the conversion between KE and PE is perfect without any dissipaton effect
• Now at x=-x₀ PE of the block is maximum ,KE is zero and the restoring force again pulls the block towards its equilibrium position.This way block would keep on oscillating
• From all discussion ,we see that total mechanical energy (KE+PE) of the system always remain constant which is equal to Kx₀²/2
• We can analyze this as follows
  x=x₀ : v=0 : KE=0 : PE=Kx₀²/2 : TE=Kx₀²/2
  x=x: v=v : KE=mv²/2=k(x₀²-x²)/2 : PE=Kx²/2 : TE=Kx₀²/2
  x=0 : v=v_max: KE=mv_max²/2=Kx₀²/2 : PE=0 : TE=Kx₀²/2
  x=-x₀: v=0 : KE=0 : PE=Kx₀²/2 : TE=Kx₀²/2
• Graph of KE and PE are shown below in figure
  
  
• From above graph we can easily conclude that when KE increases PE decreases and when PE increase KE decrease but total energy of the system remains constant
• Above discussed case is the ideal case when other dissipated forces like frictional forces or air resistance are assumed to be absent
• However in realty some part of the energy of the system is dissipated due to these resistive forces and after some time system looses all its energy and come to rest

(11) Conservative Forces

• Consider the gravitational force acting on a body .If we try to move this body upwards by applying a force on it then work is done against gravitation
• Consider a block of mass m being raised to height h vertically upwards as shown in fig 8(a) .Workdone in this case is mgh
• Now we make the block travel the path given in figure 8(b) to raise its height h above the ground.In this path workdone during the horizontal motion is zero because there is no change in height of the body due to which there would be no change in gravtitaional PE of the body and if there is no change in the speed KE would also remains same
• Thus for fig 8(b) if we add up the workdone in two vertical paths the result we get is equal to mgh
  
• Again if we move the the block to height h above the floor through an arbitary path as shown in fig 8(c) ,the workdone can be caluctlated by breaking the path into elementary horizontal and vertical portions
• Now workdone along the horizontal path would be zero and along the vertical paths its add up to mgh
• Thus we can say that workdone in raising on object against gravity is independent of the path taken and depends only on the intial and final position of the object
• Now we are in position to define the conservative forces
  " If the workdone on particle by a force is independent of how particle moves and depends only on initial and final position of the objects then such a force is called conservative force"
  Gravitatinal force,electrostatic force ,elastic force and magnetic forces are conservative forces
• Total workdone by the conservative force is zero when particle moves around any closed path returning to its initial position
• Frictional forces and viscous forces are examples of non-conservative forces as these forces always oppose the motion and result in loss in KE
• Concept of PE is associated with conservative forces only .No such PE is associated with non-conservative forces like frictional forces

(12) Power

• Power is defined as rate of doing the work
• if ΔW amount of work is done in time interval Δt ,the instantanous power delivered will be
  P=ΔW/Δt or P=dW/dt
• For total workdone W in total time t,then average power
  P_avg=W/t
• If P does not vary with time ,then P=P_avg
• SI unit of power is joule/sec also called watt
• Another wunit of power is horsepower(hp)
  1Hp=746watt

(13) Principle of conservation of energy

• We have already learned that KE+PE remains constant or conserved in absence of dissipative power
• As discussed in case of spring mass system in reality dissipative forces like friction and air resistance are always present and some of the energy of the system gets dissipated in the force of heat energy by increasing the internal energy of the spring
• This continues until system finally comes to rest
• If one can somehow measure this energy ,the sum of KE,PE and this internal (or energy dissipated) would remain constant .This can be extended to all types of energy
• Energy can not be created or destroyed .It can only be transferred from one form to another form.Total energy in a closed system always remains constant
• This is the law of conservation of energy although emprical one but it has never been found violated

Solved examples

Question 1. A body fall from height H.if t₁ is time taken for covering first half height and
t₂ be time taken for second half.Which of these relation is true for t₁ and t₂
a. t₁ > t₂
b. t₁ < t₂
c t₁=t₂
d Depends on the mass of the body

Solution 1.
Let H be the height
then
First Half
H/2=(1/2)gt₁² ----(1)
or
(1/2)gt₁²=H/2
Also v=gt₁

Second Half
H/2=vt₂+(1/2)gt₂²
or
(H/2)=gt₁t₂+(1/2)gt₂²
or
(1/2)gt₂² =(H/2)-gt₁t₂ ---(2)

From 1 and 2
(1/2)gt₂²=(1/2)gt₁² -gt₁t₂
t₂²+2t₁t₂ -t₁²=0

or t₂=[-2t₁+√(4t₁t₂ +4t₁t₂ )]/2
or t₂=[-2t₁+2t₁√8]/2
t₂=.44t₁

so t₁ > t₂

Hence a is correct


Question 2. Which of these is true of a conservative force?
a. Workdone between two points is independent of the path
b. Workdone in a closed loop is zero
c. if the workdone by the conservative is positive,its potential energy increases
d. None of the these

Solution 2. For a conservative force
Workdone between two points is independent of the path
Workdone in a closed loop is zero
And -W=U_f-U_i
So for positive work potential energy decreases
So a and b are correct


Question 3. A simple pendulum consists of a mass attached to a light string l. if the system is oscillating through small angles which of the following is true
a.The freqiency is independent of the acceleration due to gravity g
b.The period depends on the amplitude of the ocsillation
c.the period is independent of mass m
d. the period is independent of lenght l

Solution 3.Frequency =2π√(l/g)

So it is independent of the mass



Question 4. A body of mass m is dropped from a certain height.it has velocity v₁ when it is at a height h₁ above the ground.it has velocity v₂ when it is at a height h₂ above the ground.which of the following is true
a.v₁²-v₂²=2g(h₁-h₂)
b.v₁²-v₂²=2g(h₂-h₁)
c. v₁-v₂=√2g(√h₂-√h₁)
d. v₁-v₂=√2g(√h₁-√h₂)

Solution 4
Total Energy at height h₁
=(1/2)mv₁²+2gh₁

Total energy at height h₂
=(1/2)mv₂²+2gh₂

Since we know that total energy remains constant during a free fall
Total Energy at height h₁=Total energy at height h₂
or (1/2)mv₁²+2gh₁=(1/2)mv₂²+2gh₂
or v₁²-v₂²=2g(h₂-h₁)


Question 5.A pendulum has a length l.Its bob is pulled aside from its equilibrium position through any angle θ and then released.The speed of the bob when its passes through it equalibrium position
a.&radic:2gl
b. &radic:2gl(1-cosθ)
c.&radic:2glcosθ
d.&radic:2gl(1-sinθ

Solution 5 As shown in fgure,the height attained by the bob when the string subtends an angle θ is
h=l-lcosθ
or h=l(1-cosθ)
So potential energy at this point is given by
=mgh=mgl(1-cosθ)

When the bob passes through equilibrium position,this potential energy is converted into kinectic energy
if v be the velocity of the bob the KE=(1/2)mv²

Now (1/2)mv²=mgl(1-cosθ)
or v=radic:2gl(1-cosθ)


Question 6.A delivery boy wishes to launch a 2.0 kg package up an inclined plane with sufficent speed to reach the top of the incline.The plane is 3 mlong and is inclined at 20.Coefficent of friction between the package and the inclined plane is .40. what minimum intial KE must the boy suply to the package given as sin20=.342 cos20=.940
a 40 J
b. 42.2 J
c. 42.6 J
d. 45 J

Solution 6

The incline is shown in figure
If the package travels the entire length s of the incline ,the frictional force will perform work -μNs where μ is coefficient of friction and N is normal reaction.
Let h be the height of the incline plane then the gravtational potential energy of the package will increase by mgh.

Now let assume v speed be given to the package so as to reach the top
Then kinectic energy at the intial point=(1/2)mv²
Now applying work energy thoerm
K.E_f-K.E_i=Workdone by the gravitational force + workdone done by the frictional force
Now since K.E_f=0
Also Workdone by the gravitational force=-(change in gravitational potential energy)=-mgh

Therefore
-(1/2)mv²=-mgh-μNs
or (1/2)mv²=mgh+μNs

Now s=3
N=mgcosθ
h=ssinθ

Substituting all the values
(1/2)mv²=42.2J


Question 7.Chosse the correct option
a.if Workdone by the conservative force is positive then Potential energy decreases
b. Rate of change of momentum of many particles system is proportional to net external force on the system
c.The workdone by the conservative force in closed loop is zero
d. None of the above

Solution 7-
The workdone by a conservative force is equal to the negative of the potential energy.When the wokdone is positive ,the potential energy decreases.
The rate of change of total momentum of a many -particle system is proportional to the net force external to the system ;the internal forces between particles cannot change the momentum of the system.The workdone by the conservative system is zero in closed loop
Hence a,b,c are correct
Question 8.The potential energy of a certain particle is given by
U=20x²+35z³.Find the vector force on it
a. -40xi-105z²k
b. 40xi-105z²k
c.-10xi-105z²k
d 40xi+105z²k

Solution 8.
U=20x²+35z³
F=-(∂/∂x)i--(∂/∂y)j--(∂/∂z)k
or
F=-40xi-105z²k

Matrix Match type

Question 9.
Column I
a. Frictional force
b. Gravitational force
c. Electrical force
d Viscous force

Column II
P. Workdone by the force in closed loop is zero
Q. Workdone by the force in closed loop is not zero

Solution 9
Frictional force is non conservative force
Gravitational force is conservative force
Electrical force is conservative force
Viscous force is non conservative force

And for conservative force, Workdone by the force in closed loop is zero
And for non conservative force,Workdone by the force in closed loop is not zero




Question 10.Which of the following is noninertial frame of refrence
a. A train which speeding Up
b. A train with constant speed
c. A train which speeding down
d A train at rest

Question 11.what is of these is true for Projectile motion
a. Velcoity is perpendicular to acceleration at the highest point
b. Horizontal components of velocity remains constant through out the path
c. Range of the projectile is given by Horizontal component of velocity X Time of flight
d. None of the above


Question 12. A block of mass M is pulled along a horizontal friction surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is
a)Pm/m+M
b)P
c)PM/m+M
d)Pm/M-m

Question 13.A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s², the reading of the spring balance will be
(A) 24 N   
(B) 74 N
(C) 15 N   
(D) 49 N



Question 14. A particle moves in a straight line according to
x=t³-4t²+3t

Find the acceleration of the particle at displacement equal to zero
a.(-8,-2,10)
b. (-1,-2,10)
c. (8,2,10)
d. (1,2,10)

Question 14.Which of the following does not have unit as Joule?
a) Workdone
b) Kinetic Energy
c) Potential Energy
d) Force

Solution 14
Workdone Unit is joule
KE and PE also unit is Joule
Force Unit is Newton

So Solution (d)

Match the coloum
Question 15
Column A ( Physical quantity)

P)KE
Q)Potential Energy
R)Momentum
S)Mechanical Energy

Column B ( quantities it depends on)

A)Mass
B)Velocity
C)Position of the object
D)Volume

Solution 15

KE=.5mv²

PE=mgh

Momentum=mv

Mechanical Energy= KE +PE

P -> A,B
Q-> A,C
R-> A,B
S-> A,B.C

Question 16A truck driver pushes the acceleration peddle and increase it speed from v to 2v on the level Road. The mass of the Truck is M. Which of the following is true
a) PE of the truck does not change b) Final KE of the truck is 2Mv²
c) Workdone by the acceleration peddle is 1.5Mv²
d)PE of the truck becomes four times of the initial Potential energy Solution : Since it is running on level road. PE does not change Initial KE =.5Mv²
Final KE=2Mv2 Workdone by the acceleration peddle= Final KE – Initial KE=1.5Mv²

Physics 11th Impulse and Linear Momentum

Impulse and Linear Momentum

(1) Introduction

• We have already studied about the newton's laws of motion and about their application
• It becomes difficult to use Newton's law of motion as it is while studying complex problems like collision of two objects,motion of the molecules of the gas,rocket propulsion system etc
• Thus a further study of newton's law is required to find some theorem or principles which are direct consequences of Newton's law
• We have already studied one such principle which is principle of conservation of energy.Here in this chapter we will define momentum and learn about the principle of conservation of momentum .
• Thus we begin this chapter with the concept of impluse and momentum which like work and energy are developed from Newton's law of motion

(2) Impulse and momentum

• To explain the terms impulse and momentum consider a particle of mass m is moving along x-axis under the action of constant force F as shown below in the figure
  
   
• If at time t=0 ,velocity of the particle is v₀ then at any time t velocity of particle is given by the equation
  v = v₀ + at
  where a = F/m
  can be determined from the newton's second law of motion .Putting value of acceleration in above equation\
  we get
  mv = mv₀ + Ft
  or
  Ft = mv - mv₀                   -(1)
• right side of the equation Ft, is the product of force and the time during which the force acts and is known as the impluse
  Thus
  Impulse= Ft
• If a constant force acts on a body during a time from t₁ and t₂,then impulse of the force is
  I = F(t₂-t₁)                  -(2)
  Thus impulse recieved during an impact is defined as the product of the force and time interval during which it acts
• Again consider left hand side of the equation (1) which is the difference of the product of mass and velocity of the particle at two different times t=0 and t=t
• This product of mass and velocity is known as linear momentum and is represented by the symbol p. Mathematically
  p = mv                  --(3)
• physically equation (1) states that the impulse of force from time t=0 to t=t is equal to the change in linear momentum during
• If at time t₁ velocity of the particle is v₁ and at time t₂ velocity of the particle is v₂,then
  F(t₂-t₁)=mv₂-mv₁                  -(4)
• so far we have considered the case of the particle moving in a straight line i.e along x-axis and quantities involved F,v, and a were all scalars
• If we call these quantities as components of the vectors F,v and a along x-axis and generalize the definations of momentum and impulse so that the motion now is not constrained along one -direction ,Thus we got
  Impulse=I=F(t₂-t₁)                  -(5)
  Linear momentum=p=mv                  -(6)
  where
  I=I_xi+I_yj+I_zk
  F=F_xi+F_yj+F_zk
  p=p_xi+p_yj+p_zk
  v=v_xi+v_yj+v_zk
  are expressed in terms of their components along x,y and z axis and also in terms of unit vectors
• On generalizing equation (4) using respective vectors quantities we get the equation
  F(t₂-t₁) =mv₂-mv₁                  -(7)
• So far while discussing Impulse and momentum we have considered force acting on particle is constant in direction and maagnitude
• In general ,the magnitude of the force may vary with time or both the direction and magnitude may vary with time 
• Consider a particle of mass m moving in a three-dimensional space and is acted upon by the varying resultant force F. Now from newtons second law of motion we know that
  F=m(dv/dt)
  or Fdt=mdv
• If at time t₁ velocity of the particle is v₁ and at time t₂ velocity of the particle is v₂,then from above equation we have
   
• Integral on the left hand side of the equation (8) is the impulse of the force F in the time interval (t₂-t₁) and is a vector quantity,Thus
  
  Above integral can be calculated easily if the Force F is some known function of time t i.e.,
  F=F(t)
• Integral on the right side is when evaluated gives the product of the mass of the particle and change in the velocity of the partcile
   
• using equation (9) and (10) to rewrite the equation (8) we get
   
• Equivalent equations of equation (11) for particle moving in space are
   
• Thus we conclude that impulse of force F during the time interval t₂-t₁ is equal to the change in the linear momentum of the body on which its acts
• SI units of impulse is Ns or Kgms_-1

(3) Conservation of Linear momentum

• Law of conservation of linear momentum is a extremely important consequence of Newton's third law of motion in combination with the second law of motion
• Consider two particles of mass m₁ and m₂ interacting with each other and forces acting on these particles are only the ones they exert on each other.
• Let F₁₂ be the force exerted by the particle 2 on particle 1 having mass m₁ and velocity v₁ and F₂₁=-F₁₂ be the force exerted by the particle 1 on particle 2 having mass m₂ and velocity v₂
• Applying newton second law of each particle on each partcile
  
  F₁₂=m₁(dv₁/dt)
  and F₂₁=m₂(dv₂/dt)
• from newton's third law of motion
  F₂₁=-F₁₂
  or m₁(dv₁/dt) + m₂(dv₂/dt)=0
  Since mass of the particles are not varying with time so we can write
  (d/dt)(m₁v₁ +m₂v₂)=0
  or m₁v₁ +m₂v₂=constant                 --(13)
• we have already defined the quantity mv as the momentum of the particle
• Thus we conclude that when two particles are subjected only to their mutual interactions ,the sum of the momentums of the bodies remains constant in time or we can say the total momentum of the two particles does not change becuase of the any mutual interactions between them
• For any kind of force between two particles then sum of the momentum ,both before and after the action of force should be equal i.e total momentum remains constant
• We thus arrive to the statment of principle of conservation of linear momentum
  " when no resultant external force acts on system ,the total momentum of the system remains constant in magnitude and direction" 
• In absence of external forces for a number of interacting particles,law of conservation of linear momentum can be expressed as
  m₁v₁ +m₂v₂+m₃v₃+m₄v₄+...=constant 
• Law of conservation of linear momentum is one of the most fundamental and important principle of mechanics
• This principle also holds true even if the forces between the interacting particles is not conservative
• Once again ,the total momentum of two or any number of particles of interacting particles is constant if they are isolated
  from outside influences (or no resultant external forces is acting on the particles)

(4) Recoil of a gun

• Consider the gun and bullet in its barrel as an isolated system
• In the begining when bullet is not fired both the gun and bullet are at rest.So the momentum of the before firing is zero
  p_i=0
• Now when the bullet is fired ,it moves in the forward direction and gun recoil back in the opposite direction
• Let m_b be the mass and v_b of velocity of the bullet And m_g and v_g be the velcoity of the gun after the firing
• Total momentum of the system after the firing would be
  p_f=m_bv_b +m_gv_g
• since no external force are acting on the system,we can apply the law of conservation of linear momentum to the system
  Therfore
  p_f=p_i
  or m_bv_b +m_gv_g=0
  or v_g=-(m_bv_b/m_g)
  
• The negative sign in above equation shows that velocity of the recoil of gun is opposite to the velocity of the bullet
• Since mass of the gun is very large as compared to the mass of the bullet,the velocity of the recoil is very small as compared to the velocity of the bullet

(5) Motion of the system with varying mass(Rocket)

• Uptill now while studying classical mechanics we have always considered the particle under consideration to have constant mas
• Some times it is required to deal with the particles or system of particles in which mass is varying and motion of the rocket is one such examples
• In a rocket fuel is burned and the exhaust gas is expelled out from the rear of the rocket
• The force exerted by the exhaust gas on the rocket is equal and opposite to the force exerted by the rocket to expell it
• This force exerted by the exhaust gas on the rocket propels the rocket forwards
• The more gass is ejected from the rocket ,the mass of the rcoket decreases
  
   
• To analyze this process let us consider a rocket being fired in upwards direction and we neglect the resistance offered by the air to the motion of the rocket and variation in the value of the acceleration due to gravity with height
• Figure above shows a rocket of mass m at a time t after its take off moving with velocity v.Thus at time t momentum of the rocket is equal to mv.THus
  p_i=mv
• Now after a short interval of time dt,gas of total mass dm is ejected from the rocket
• If v_g represents the downward speed of the gas relative to the rocket then velocity of the gas relative to earth is
  v_ge=v-v_g
  And its momentum equal to
  dmv_ge=dm(v-v_g)
• At time t+dt,the rocket and unburned fuel has mass m-dm and its moves with the speed v+dv.Thus momentum of thee rocket is
  =(m-dm)(v+dv)
  
• Total momentum of the system at time t+dt is
  p_f=dm(v-v_g)+(m-dm)(v+dv)
  Here system constitute the ejected gas and rocket at the time t+dt
• From Impulse momentum relation we know that change in momentum of the system is equal to the product of the resultant external force acting on the system and the time interval during which the force acts
• Here external force on the rocket is weight -mg of the rocket ( the upward direction is taken as positive)
• Now
  Impulse=change in momentum
  F_extdt=p_f-p_i
  or
  -mgdt=dm(v-v_g)+(m-dm)(v+dv) - mv
  or
  -mgdt=mdv-v_gdm-dmdv
  term dmdv can be dropped as this product is neglibigle in comparison of other two terms
  Thus we have
   
• In equation (14) dv/dt represent the acceleration of the rocket ,so mdv/dt =resulant force on the rocket
  
  Therefore
  Resultant Force on rocket=Upthrust on the rocket - weight of the rocket
  where upthrust on rocket=v_g (dm/dt)
• The upthrust on rocket is proportional to both the relative velocity (v_g) of the ejected gas and the mass of the gas ejected per unit time (dm/dt)
• Again from equation (14)
  
  As rocket goes higher and higher ,value of the acceleration due to gravity g decreases continously .The values of v_g and dm/dt parctically remains constant while fuel is being consumed but remaining mass m decreases continously .This result in continous increase in acceleration of rocket untill all the fuel is burned up
• Now we will find the relation between the velocity at any time t and remaining mass.Again from equation (15) we have
  dv=v_g (dm/m) -gdt
• Here dm is a +ve quantity representing mass ejected in time dt.So change in mass of the rocket in time dt is -dm.So while calculating total mass change in rocket,we must change the sign of the term containing dm
  dv=-v_g (dm/m) -gdt                 --(16)
• Initially at time t=0 if the mass and velocity of the rocket are m₀ and v₀ respectively.After time t if m and v are mass and velocity of the rocket then integrating equation (16) within these limits
  
  On evaluating this integral we get
  v-v₀=-v_g(ln m- ln m₀)-g(t-0)
  or v=v₀+v_gln(m₀/m) -gt                (17)
• equation(17) gives the change in velocity of the rocket in terms of exhaust speed and ration of initial amd final masses at any time t
• The speed acquired by the rocket when the whole of the fuel is burned out is called burn-out speed of the rocket

Solved examples

Question 1 .A 1 kg ball moving at 12 m/s collides head on with 2 kg ball moving with 24 m/s in opposite direction.What are the velocities after collision if e=2/3?
a. v₁=-28 m/s,v₂=-4 m/s
b. v₁=-4 m/s,v₂=-28 m/s
c. v₁=28 m/s,v₂=4 m/s
d. v₁=4 m/s,v₂=28 m/s

Solution 1

Let v₁ and v₂ be the final velocities of the mass

Since the linear momentum is conserved in the collision
Momentum before =Momentum after
1*12+2*-24=1*v₁+2*v₂
Which becomes
-36=v₁+2v₂ ----1

Now
e=(v₂ -v₁)/(u₁ -u₂)

or 2/3= (v₂ -v₁)/[12-(-24)]
or
v₂ -v₁=24 ----2

Solving 1 and 2

v₂=-4
v₁=-28

Hence a is correct


Question 2.A moving bullet hits a solid target resting on a frictionless surface and get embeded in it.What is conserved in it?
a. Momentum Alone
b KE alone
c. Both Momentum and KE
d. Neither KE nor momentum

Solution 2 Since no external force is present,Momentum is conserved in the collision
Since the collison is in elastic ,KE is not conserved




Question 3. A stationary body of mass 3 kg explodes into three equal parts.Two of the pieces fly off at right angles to each other with the velocities 2i m/s and 3j m/s.If the explosion takes place in 10^-3 sec.find out the average force action on the third piece in N
a.(-2i-3j)10³
b. (2i+3j)10³
c (2i-3j)10^-3
d. none of these

Solution 3.

Net momentum before explosion zero
Since momentum is conserved in explosion
Net momentum after collosion is zero

Momentum of first part after explosion=2i
Momentum of second part after explosion=3j

So momentum of third part after explosion=-(2i+3j) as net momentum is zero

Now Net change is momentum of this part =-(2i+3j)
Now we know that
Average force X time =Net change in momentum
Average force=-(2i+3j) 10³

hence a is correct



Question 4.A bullet of mass m is fired horizontally with a velocity u on a wooden block of Mass M suspended from a support and get embeded into it.The KE of th wooden + block system after the collisson
a.m²u²/2(M+m)
b.mu²/2
c. (m+M)u²/2
d. mMu²/2(M+m)

b>Solution 4.

Intial velocity of bullet=u
Intial velocity of block=0
So net momentum before collison=mu

Let v be the velocity after collision
Then Net momentum after collision=(M+m)v

Now linear momentum is conserved in this collision
so
mu=(M+m)v
or v=mu/(M+m)

So kinetic energy after collision
=(1/2)m²u²/2(M+m)

Hence a is correct


Question 5.A body of Mass M and having momentum p is moving on rough horizontal surface.If it is stopped in distance s.Find the value of coefficient of friction
a.p²/2M²gs
b. p/2Mgs
c. p²/2Mgs
d. p/2M²gs

Solution 5.

Deceleration due to friction=μg

Intial velocity=P/M

Now v²=u² -2as
as v=0
P²/M²=2μgs
or μ=P²/2gsM²

Hence a is correct




Question 6.A rockets works on the principle of conservation of
a. Linear momentum
b.mass
c.energy
d. angular momentum

Solution 6. A rocket works on the principle of linear momentum.



Question 7.A flat car of weight W roll without resistance along on a horizontal track .Intially the car together with weight w is moving to the right with speed v.What invcrement of the velocity car will obtain if man runs with speed u reltaive to the floor of the car and jumps of at the left?
a.wu/w+W
b. Wu/W+w
c. (W+w)u/w
d. none of the above

Solution 7 Considering velocities to the right as positive
The intial momentum of the system is
=[(W+w)/g]v

Let Δv be the increment in velocity then

Final momentun of the car is
(W/g)(v+Δv)

While that of man is
(w/g)(v+Δv-v)

Since no external forces act on the system,the law of conservation of momentum gives then
[(W+w)/g]v=(W/g)(v+Δv)+(w/g)(v+Δv-v
or Δv=wu/(W+w)



Question 8.Consider the following two statements.
STATEMENT 1 Linear momentum of a system of particles is zero.
STATEMENT 2 Kinetic energy of system of particles is zero.
(A) A does not imply B and B does not imply A.
(B) A implies B but B does not imply A
(C) A does not imply B but b implies A’  
(D) A implies B and B implies A.

Solution 8
Net momentum=m₁v₁+m₂v₂
Net Kinectic Energy=(1/2)m₁v₁²+(1/2)m₂v₂²

Let v₁=v ,v₂=-v and m₁=m₂=m
Then Net momentum=0 but Net Kinectic Energy is not equal to zero

Now lets v₁= v₂=0

Then Net Kinectic Energy=0 and Net momentum=0
Hence (c) is correct